Open image of surjective local diffeomorphism

Question

Let $M,N$ be smooth manifolds (in $\mathbb{R}^n$), and let $f: M\rightarrow N$ be a local diffeomorphism.

Is it true that $f(M)$ is open in $N$?

Fix $q \in f(M)$: we have $p \in M$ such that $f(p)=q$. Since $f$ is a local diffeomorphism we can find $U \subseteq N$ open such that $f(U)$ is open, and it is an open set in $N$ around $q$ contained in $f(M)$. Hence $f(M)$ is open.

Is this correct?

No. I mean $f(M)$ automatically open since the codomain of the map is $N$, which is by definition of topology open in itself. — Kelvin Lois, Apr 28 '20 at 12:53
@Sou You're right, I confused the hypothesis and I've changed the question — Lilla, Apr 28 '20 at 14:19
This can be treated as a particular case of https://math.stackexchange.com/questions/1826878/show-that-every-local-homeomorphism-is-continuous-and-open-therefore-bijective-l — Paweł Czyż, Apr 28 '20 at 14:31

score 1 · Answer 1 · answered Apr 28 '20 at 11:50

Since $f$ is a local diffeomorphism there exists an open neighbourhood $U$ of $q$ such that $f:U\rightarrow V$ is a diffeomorphism, where $V=f(U)$. This implies that $V$ is an open neighbourhood of $p$ contained in $N=f(M)$ by surjectivity of $f$.

You can proceed in this way for any point $p\in N$, i.e. in $f(M)$, and hence by surjectivity you get that $f(U)$ is open.

Open image of surjective local diffeomorphism

1 Answers1